An integral calculus I know that $\displaystyle \frac{1}{2i\pi}\int_0^{2\pi}\frac{ib \cos t - a \sin t}{a\cos t+ib\sin t} \, dt =1.$
I'm trying to use this to find the value of $\displaystyle \int_0^{2\pi}\frac{dt}{a^2\cos^2t+b^2\sin^2t}.$
Is it possible? Any hint would be much appreciated.
 A: Note that we can write
$$\begin{align}
\frac{ib\cos(t)-a\sin(t)}{a\cos(t)+ib\sin(t)}&=\frac{iab }{a^2\cos^2(t)+b^2\sin^2(t)}+\frac{\frac12(b^2-a^2)\sin(2t) }{a^2\cos^2(t)+b^2\sin^2(t)}\end{align} \tag 1$$
Therefore, integrating $(1)$ reveals 
$$\frac{1}{2\pi i}\int_0^{2\pi}\frac{ib\cos(t)-a\sin(t)}{a\cos(t)+ib\sin(t)}\,dt=\frac{ab}{2\pi}\int_0^{2\pi}\frac{1 }{a^2\cos^2(t)+b^2\sin^2(t)}\,dt \tag 2$$
since 
$$\begin{align}
\int_0^{2\pi}\frac{\frac12(b^2-a^2)\sin(2t) }{a^2\cos^2(t)+b^2\sin^2(t)}\,dt&=\int_{-\pi}^{\pi}\frac{\frac12(b^2-a^2)\sin(2t) }{a^2\cos^2(t)+b^2\sin^2(t)}\,dt \tag 3\\\\
&=0 
\end{align}$$
due to the odd symmetry and $2\pi$-periodicity of the integrand in $(3)$!
Therefore, rearranging $(2)$ and using $\frac{1}{2\pi i}\int_0^{2\pi}\frac{ib\cos(t)-a\sin(t)}{a\cos(t)+ib\sin(t)}\,dt=1$ yields
$$\bbox[5px,border:2px solid #C0A000]{\int_0^{2\pi}\frac{1 }{a^2\cos^2(t)+b^2\sin^2(t)}\,dt=\frac{2\pi}{ab}}$$ 
A: You can multiply and divide the integral that you've mentioned
 by $a \cos t -ib \sin t$ as follows,
$$  \frac{1}{2i\pi}\int_0^{2\pi}\frac{(ib \cos t - a \sin t)(a \cos t - ib \sin t)}{(a\cos t+ib\sin t)(a \cos t - ib \sin t)} \, dt =1 $$
$$ \therefore \frac{1}{2i\pi}\int_0^{2\pi}\frac{aib \cos^2 t + b^2 \sin t \cos t - a^2 \sin t \cos t + aib \sin^2 t}{(a\cos t+ib\sin t)(a \cos t - ib \sin t)} \, dt =1 $$
$$ \therefore \frac{1}{2i\pi}\int_0^{2\pi}\frac{aib + \frac{(b^2 - a^2)}{2}\sin {2t}}{a^2 \cos^2 t + b^2 \sin^2 t} \, dt =1 $$
You can then use the above simplification to compute the desired integral.
